Tuyển tập đề thi vô địch bất đẳng thức thế giới P2

Tuyển tập đề thi vô địch bất đẳng thức thế giới P2 , tài liệu tham khảo, tài liệu gồm các bài toán bất đẳng thức cực khó, các bạn có đào sâu kiến thức toán về mảng này, Tai liệu được viết bằng tiêng anh. Chúc các bạn học tốt. | Old and New Inequalities 31 When do we have equality Gazeta Matematicä Solution First we write the inequality in the following form A 8w A 9z A 6 4 1 3rc 1 1 1 - 74. x j y J z J But this follows immediately from Huygens Inequality. We have equality for 3 X 2 y - z 1. 11. Mihai Piticari Dan Popescu Prove that 5 a2 A A 6 a3 b3 c3 1 for all a b c 0 with a b c 1. Solution Because a b c 1 we have a3 b3 c3 3abc a2 b2 c2 ab be ca. The inequality becomes 5 a2 b2 c2 18abc 6 a2 b2 c2 6 ab be ca 1 O O 18abc 1 2 ab bc ca 1 6 a be ca O 8 ab bc ca 2 18abc 4 a be ca 1 9abc iv 1 2a l 2 1 2c abc O b c a c a b a b c abc which is equivalent to Schur s Inequality. 12. Mircea Lascu Let Xi aq xn E R n 2 and a 0 a 2 a and x xi . x2 -. Prove that Xi E n - 1 such that xi 0 for all n i G 1 2 . n . Solution Using the Cauchy-Schwarz Inequality we get Thus a2 2axi x2 a2 n l x2 aq aq--------I 0 and the conclusion follows. 32 Solutions 13. Adrian Zahariuc Prove that for any a b c G 1 2 the following inequality holds Solution The fact that a b c G 1 2 makes all denominators positive. Then by a b a b c y a 4by c cy a the last one coming from a b 2 fab and b c 2 5c. Writing the other two inequalities and adding them up give the desired result. 14. For positive real numbers a b c such that abc 1 prove that a b c 7 G - G - a b c. b c a First solution If ab be ca a b c then the Cauchy-Schwarz Inequality solves the problem a b c 2 t ï r - - - b c a abc Otherwise the same inequality gives -----7------ a b c. lab be ca 2 b c a abc a b c- a b c here we have used the fact that abc 1 . Second solution Replacing a b. c by ta tb tc with t the value of the quantity in abc the left-hand side of the inequality and increases the value of the right-hand side and makes at bt ct abet3 1. Hence we may assume without loss of generality that y z x abc 1. Then there exist positive real numbers x y z such that a b c The Rearrangement Inequality gives x y z x3 y3 z3 x2y y2z z2x. Old and New Inequalities 33 .